So sánh, n ∈ N $\frac{n}{n+1}$ và $\frac{n+1}{n+2}$ 06/12/2021 Bởi Athena So sánh, n ∈ N $\frac{n}{n+1}$ và $\frac{n+1}{n+2}$
$\frac{n}{n+1}=$ $\frac{n(n+2)}{(n+1)(n+2)}=\frac{n^2+2n}{(n+1)(n+2)}$ (1) $\frac{n+1}{n+2}=$ $\frac{(n+1)^2}{(n+1)(n+2)}=$ $\frac{n^2+2n+1}{(n+1)(n+2)}$ (2) Từ (1) vào (2) ⇒ $n/(n+1)<(n+1)(n+2)$ Bình luận
có n/n+1=n+1-1/n+1=1-(1/n+1) n+2/n+3=n+3-1/n+3=1-(1/n+3) Vì 1-(1/n+1)<1-(1/n+3) hay n/n+1 < n+2/n+3 Bình luận
$\frac{n}{n+1}=$ $\frac{n(n+2)}{(n+1)(n+2)}=\frac{n^2+2n}{(n+1)(n+2)}$ (1)
$\frac{n+1}{n+2}=$ $\frac{(n+1)^2}{(n+1)(n+2)}=$ $\frac{n^2+2n+1}{(n+1)(n+2)}$ (2)
Từ (1) vào (2) ⇒ $n/(n+1)<(n+1)(n+2)$
có n/n+1=n+1-1/n+1=1-(1/n+1)
n+2/n+3=n+3-1/n+3=1-(1/n+3)
Vì 1-(1/n+1)<1-(1/n+3) hay n/n+1 < n+2/n+3